Question:

Find the derivative of the following functions from first principle:
(i) −x (ii) (-x)-1 (iii) sin (x + 1) (iv) cos (x – π8\frac{\pi}{8})

Updated On: Nov 1, 2023
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Solution and Explanation

(i) Let f(x) = -x. Accordingly, f(x+h)=-(x+h)
By first principle,
f'(x)=limh0\lim_{h\rightarrow 0} f(x+h)f(x)h\frac{f(x+h)-f(x)}{h}
=limh0\lim_{h\rightarrow 0} (x+h)(x)h\frac{-(x+h)-(-x)}{h}
=limh0\lim_{h\rightarrow 0} xh+xh\frac{-x-h+x}{h}
=limh0\lim_{h\rightarrow 0} hh\frac{-h}{h}
=limh0\lim_{h\rightarrow 0}(-1)=-1


(ii) Let f(x)=(-x)-11x\frac{1}{-x} = 1x-\frac{1}{x} Accordingly, f(x+h)=1x+h-\frac{1}{x+h}
By first principle,
f'(x)=limh0\lim_{h\rightarrow 0} f(x+h)f(x)h\frac{f(x+h)-f(x)}{h}
=limh0\lim_{h\rightarrow 0} 1h\frac{1}{h}[1x+h(1x)-\frac{1}{x+h}-(\frac{-1}{x})]
=limh0\lim_{h\rightarrow 0}1h\frac{1}{h}[1x+h+1x\frac{-1}{x+h}+\frac{1}{x}]
=limh0\lim_{h\rightarrow 0}1h\frac{1}{h}[x+(x+h)x(x+h)\frac{-x+(x+h)}{x(x+h)}]
=limh0\lim_{h\rightarrow 0}1h\frac{1}{h}[x+x+hx(x+h)\frac{-x+x+h}{x(x+h)}]
=limh0\lim_{h\rightarrow 0}1h\frac{1}{h}[hx\frac{h}{x}(x+h)]
=limh0\lim_{h\rightarrow 0} 1x\frac{1}{x}(x+h)
1x\frac{1}{x}.x - 1x2\frac{1}{x^2}


(iii) Let f(x) = sin (x + 1). Accordingly, f(x+h)=sin(x+h +1)
By first principle,
f'(x)=limh0\lim_{h\rightarrow 0} f(x+h)f(x)h\frac{f(x+h)-f(x)}{h}
=limh0\lim_{h\rightarrow 0} 1h\frac{1}{h}[sin(x+h+1) -sin(x+1)]
=limh0\lim_{h\rightarrow 0} 11h\frac{1}{h}[2cos(x+h+1+x+12\frac{1}{2}) sin(x+h+1-x-1)]
=limh0\lim_{h\rightarrow 0} 1h\frac{1}{h}[ 2cos (2x + h+22\frac{2}{2}) sin(h2\frac{h}{2})]
=limh0\lim_{h\rightarrow 0} 1h\frac{1}{h}[ 2cos (2x+h+22\frac{2x+h+2}{2}) sin(h2\frac{h}{2})/(h2\frac{h}{2})]
=cos(2x+0+22\frac{2x+0+2}{2}).1 [limh0\lim_{h\rightarrow 0} sinxx\frac{sin\,x}{x} =1]
=cos(x+1)


(iv) letf(x) cos(x-π8\frac{\pi}{8}). Accordingly, f(x+h)=cos(x+h -π8\frac{\pi}{8})
By first principle,
f'(x)=limh0\lim_{h\rightarrow 0} f(x+h)f(x)h\frac{f(x+h)-f(x)}{h}
=limh0\lim_{h\rightarrow 0}1h\frac{1}{h}[cos(x+h-π8\frac{\pi}{8})- cos(x-π8\frac{\pi}{8})]
=limh0\lim_{h\rightarrow 0}1h\frac{1}{h}[2sin(x+hπ8+π8)2sin(x+hπ8x+π82)\frac{-2sin(x+h-\frac{\pi}{8}+-\frac{\pi}{8})}{2\,sin(\frac{x+h-\frac{\pi}{8}-x+\frac{\pi}{8}}{2})}]
=limh0\lim_{h\rightarrow 0}1h\frac{1}{h}[2sin(2x+hπ4)2\frac{-2sin(2x+h-\frac{\pi}{4})}{2} sinh2sin \frac{h}{2}]
=limh0\lim_{h\rightarrow 0}[-sin(2x+hπ42\frac{2x+h-\frac{\pi}{4}}{2}sinh2h2\frac{sin\frac{h}{2}}{\frac{h}{2}}]
=-sin(2x+0π42\frac{2x+0-\frac{\pi}{4}}{2}) .1
=-sin(x-π8\frac{\pi}{8})

 

 

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